What are the intervals on the number line for |x| < 2 and |x| > 0?

[mathdad]

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show each interval on the number line.

1. | x | < 2

Solution:

-2 < x < 2

This can be written as (-2, 2).

----(-2-----0----2)----

Correct?

2. | x | > 0

Solution:

The point x lies to the right 0.

----(0--------------->

Correct?

 

[Evgeny.Makarov]

1. | x | < 2

Solution:

-2 < x < 2

This can be written as (-2, 2).

----(-2-----0----2)----

Correct?

Yes.

2. | x | > 0

Solution:

The point x lies to the right 0.

----(0--------------->

Correct?

What about $x=-1$?

 

[mathdad]

Yes.

What about $x=-1$?

What do you mean x = -1?

How does | x | > 0 involve -1?

 

[Evgeny.Makarov]

You are claiming that $|x| > 0$ iff the point $x$ lies to the right 0 for all $x$. I suggest you check this claim for $x=-1$.

If a person says, "All birds can fly" and another asks, "What about ostrich?", this means that the second person questions the original statement and offers a potential counterexample.

 

[mathdad]

You are claiming that $|x| > 0$ iff the point $x$ lies to the right 0 for all $x$. I suggest you check this claim for $x=-1$.

If a person says, "All birds can fly" and another asks, "What about ostrich?", this means that the second person questions the original statement and offers a potential counterexample.

Can you demonstrate why x = -1 is needed to be checked? Show me what you're talking about.

 

[Evgeny.Makarov]

Can you demonstrate why x = -1 is needed to be checked?

No, I prefer you check first whether you claim is true for $x=-1$.

Show me what you're talking about.

We are talking about your claim: $|x|>0\iff x\text{ lies to the right of 0}$ for all $x$. To check a statement of the form $\forall x.\,(P(x)\iff Q(x))$ for a particular $x_0$ (here $\forall$ is read "for all") means that you determine whether $P(x_0)$ is true, whether $Q(x_0)$ is true and then whether these answers coincide. If both are true or false, then the statement is true for that particular $x_0$. If one is true and the other is false, the equivalence is false.

In general it's a bad idea to ask someone "Why do I have to do this?". Legitimate questions in math are usually in the form "Is this statement true?" and "Why is this true?". For example, if you are reading a proof, it's not good to say "Why does the author choose this path to prove it?", but it's OK to ask "Why does this statement imply the following?". If someone gives you an object that may happen to be a counterexample to your statement, it's not good to ask, "How did you come up with it?" or "Why do I have to check it?". You should primarily be interested in whether your statement is true or false. Other questions may be legitimate, but they are not the first ones.

 

[mathdad]

No, I prefer you check first whether you claim is true for $x=-1$.

We are talking about your claim: $|x|>0\iff x\text{ lies to the right of 0}$ for all $x$. To check a statement of the form $\forall x.\,(P(x)\iff Q(x))$ for a particular $x_0$ (here $\forall$ is read "for all") means that you determine whether $P(x_0)$ is true, whether $Q(x_0)$ is true and then whether these answers coincide. If both are true or false, then the statement is true for that particular $x_0$. If one is true and the other is false, the equivalence is false.

In general it's a bad idea to ask someone "Why do I have to do this?". Legitimate questions in math are usually in the form "Is this statement true?" and "Why is this true?". For example, if you are reading a proof, it's not good to say "Why does the author choose this path to prove it?", but it's OK to ask "Why does this statement imply the following?". If someone gives you an object that may happen to be a counterexample to your statement, it's not good to ask, "How did you come up with it?" or "Why do I have to check it?". You should primarily be interested in whether your statement is true or false. Other questions may be legitimate, but they are not the first ones.

If x = -1, then |-1| = 1 > 0.

 

[MarkFL]

When I see an inequality of the form:

$$|x-h|>k$$

I read that as "x is the set of all points on the number line whose distance from h is greater than k." So I plot the point h, then move k units to the left and plot and open circle and shade to the left with an arrow pointing left, and then I move k units to the right of h and plot an open circle and shade to the right with and arrow pointing right to graph the solution.

Doing so for this problem yields:

\begin{tikzpicture}[scale=2.5]
\path [draw=black, fill=white, thick] (0,0) circle (2pt);
\draw[latex-latex] (-2.5,0) -- (2.5,0) ;
\draw[->,thick] (0,0) -- (2.25,0);
\draw[->,thick] (0,0) -- (-2.25,0);
\foreach \x in {-2,-1,0,1,2}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {-2,-1,0,1,2}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

We see that the solution is that x is all real numbers except 0. In interval notation, that would be:

$$(-\infty,0)\,\cup\,(0,\infty)$$

In set-builder notation, it would be:

$$\{x|x\ne0\}$$

This is read "x such that x is not equal to zero." :D

 

[mathdad]

Very good.

 

[Evgeny.Makarov]

We are talking about your claim: $|x|>0\iff x\text{ lies to the right of 0}$ for all $x$. To check a statement of the form $\forall x.\,(P(x)\iff Q(x))$ for a particular $x_0$ (here $\forall$ is read "for all") means that you determine whether $P(x_0)$ is true, whether $Q(x_0)$ is true and then whether these answers coincide. If both are true or false, then the statement is true for that particular $x_0$. If one is true and the other is false, the equivalence is false.

If x = -1, then |-1| = 1 > 0.

So we are talking about the claim
\[
|x|>0\iff x\text{ lies to the right of 0}
\]
for all $x$. I suggested checking it for $x=-1$. You determined that $|x|>0$ is true because $\lvert-1\rvert=1>0$. However, you did not check whether the right-hand side is true, namely, whether $-1$ lies to the right of $0$. Obviously, it is false. Since the truth values of the left- and right-hand side of the equivalence are different (true and false, respectively), the equivalence is false. Therefore, it is not true that $|x|>0\iff x\text{ lies to the right of 0}$ holds for all $x$: it does not hold for $-1$.

The difficulty may come from not being clear on what the problems asks. The problem asks to draw the set $A$ on the number line such that $|x|>0\iff x\in A$ holds for all $x$. The symbol $\in$ means "is an element of". For special forms of inequalities in the left-hand side you may have methods of solving this problem, like the method described by Mark. However, the left-hand side can be anything: an equation, an inequality, a statement written with words, etc. It is important to understand what it means to solve an equation or an inequality, and for this you need to understand the concepts that are part of "$|x|>0\iff x\in A$ holds for all $x$". These concepts are: equivalence, set, element, universal quantification ("for all").

 

[mathdad]

Thank you everyone.